 So, since we've been at technology failure today, so I won't use that board because the camera doesn't work. So, use it anyway? What, you want to go up and man the camera? I'd like people to explain that I go over here, you know I love the camera, I can do whatever I want to do here. So I'll just use these two boards. Okay, so, I would like to finish with the Taylor series stuff so that we can do one, two other things. So at the end of the last class, so one of the things, so let me remind you, we have a power series, so we have a Taylor, a Lauren series, so a Taylor series looks like a power series, a power series like this, where we're not necessarily centered at zero, and the Lauren series is a special case of a Taylor series. Let me actually not write it as A, N. You write it since it's a Taylor series and not a power series. So this is f prime of some number just to the nth derivative, some number n factorial x minus c g n, and then the Lauren series is the same thing except c is zero, nth derivative is zero, n factorial x to the n. So when we have an infinite series like this, as one of the main utilities of series, it's often difficult to add up infinitely many things, so we're going to stop somewhere. So the question is, which I wrote at the end of the last time, so both of these define some function, and this gives us f of x. So the question is how good is the approximation if we stop after n terms? Write some number which is called the remainder, which is the difference between the actual summing up to infinity minus when we stop it in terms. The difference here is we want something that's good for all the x's. I want to think of this as a function of a polynomial of degree n. This will be a polynomial of degree big n, and we want to know how far off will that polynomial be from the actual function. So we can't quite use just the integral test or, you know, those two series, those two remainder things that we had before where we integrate the tail or it's an alternating series and we use the alternating series remainder. But it's very similar, and the theorem is that, so the answer is if we know that the n derivative is no bigger than some number for all the x's that we care about, then the remainder is less than or equal to basically the next term but where we replace the value of the derivative by this value. So it's no bigger than m over two degrees, this big mess. Times, these are n plus 1's, I'm sorry. So it's just like the alternating series test except we have to plug in a value and the place where we replace the coefficient with is not necessarily the place where we're evaluating. I don't know if you quite followed that. So let me do an example. Right? So does this make sense to people kind of? Not at all? Not at all. Okay, so let me do an example. Let me first start with one we know. One to use e to the x is about one plus x plus n squared over two, and I'm just going to stop here at two. So how good is this? So what's the worst approximation if I take say x between minus point one and plus one? In other words, how far off will this go? This is the biggest error I made. Right? The graph of e to the x looks like this and this polynomial is a parabola. So how good is this parabola? How closely does this parabola fit if I go out to point one on the other side? What's this thing? So we have to do a little bit of work. We have to see what the third derivative is at the center. So here my center is c equals zero. I'm moving a Taylor series at zero. And I want to know what's the biggest error if I'm going to replace the calculation of e by calculation of this degree to a polynomial? How far off will I be? And so this theorem tells me, well I've got to figure out what's the biggest, the third derivative is in this region. And then I'm okay. So I need the maximum of the third derivative for x in this range. That will give me my m. So here the third derivative I probably shouldn't know. That's okay. So f of x is e to the x. At prime, this is a stupid example maybe I shouldn't have done a minus x. Can I use a minus x? It doesn't matter. f prime of x is e to the x. f double prime of x, you know, look at the end. f triple prime of x is e to the x. So what's the biggest e to the x is between point one and minus point one. e to the point one. Okay? So that's some number which unfortunately I don't know what to talk about in my head. If somebody has a calculator maybe you could probably afford it. And so now that tells me, so I guess I'll do over here, that my error is no more than e to the point one over three factorial times as long as I say within point one. My answer will be good to this much. Does anyone have a calculator tell me what e to the point one is? I mean I can guess. What? 1.105. So 1.105 is more digits, right? So my error will be no more than this. And remember x here is less than point one in absolute value. So this is really point one cubed. So this is, so that's point, point one cubed will give me point zero zero one which will move this back three places. Probably divide that by six. This is point zero zero zero. So this is good to within point zero zero within better than three places if we replace e by one plus x plus x square as long as we keep the value small. Then e is just like this thing. We're better than three places after this. Okay? Of course, yeah. If I stopped at the second derivative I would be estimating how close e is to one plus x, right? I mean since all the derivatives of e are e to the x this was easy. So could I have stopped at the second derivative? Well it wouldn't be estimating what I wanted. In this case the m is the same for all of them because e to the x is its own derivative. I'm trying to say the derivative is no bigger than some number. So let me do another example where the function changes when I take it. I'm saying I want to use this variation for this range of axis. The biggest at the third derivative is when it's point one. But it's never point one. But it's as close to point one as I want. So what could I use? Point nine nine nine nine nine nine point oh nine nine nine nine nine nine nine nine nine. I get higher than nine. So I have to use it. So I could also use 10, right. my answer is better than. So if I use the hundreds, okay, it's fine. It would give me a terrible estimation. You know, I'm saying that this distance is less than this distance. It's true. It's fine. It's just stupid. You know, if I wanted to estimate my height, oh, I'm shorter than the ceiling. Okay? So, I use point one because it's what I had at hand. So for example, if I'm doing one with the sine, I could use one. I could also use, you know, if I'm estimating the sine between zero and point one, I could use the sine of point one, but maybe I don't know the sine of point one. But the sine's less than one. It will give me an answer. It won't be the best answer, but it will be better than nothing. And it won't be wrong. It will just be not as precise as it could have been. So, see, if he didn't have a calculator, I wouldn't have used one point one. I would have just guessed what he did to point one is by sort of messing with it and said, oh, it's less than two. But I would have put a two here. It would have still given me not too bad an estimate. If I put a two here, and I had a third point oh oh one, which is point oh oh three. Okay? So what? It's close enough. Okay, so let me do one more example of this and then move on to something else. So suppose I want to use Taylor series to approximate, I don't know, a few groups, near equals eight. And let's just, okay, six contemporary. That's two on either side. Okay? So same question in some sense. Just a different function. So here, I don't need to know the series. So there's my function, x to the one third. And I want to know, I'm going to start, I'm just going to derive the Taylor series. And I want it near x equals eight. And I know the cube root of eight. That's two. Okay? So I need to calculate several derivatives of this. I'm doing n equals four. So this is one third x to the minus two thirds. I don't really feel I will need this. f prime of eight is one third x to the eight to the minus two thirds. So cube root of eight is two. Two to the minus two is four. Second derivative is minus two ninths x to the minus, why can't I do this? Five thirds, which is an eight. This is minus two ninths. One over two to the fifth is 32, right? Why can't I do that? Which is four 32. Maybe I should do n equals three, I'm getting tired. Okay, so this is ten twenty sevenths x to the minus eight thirds. This is minus, this is plus ten twenty sevenths one over two to the eighth is some number that I've forgotten. One more. Fourth derivative is minus eighty over eighty one. Let's see, right? X to the minus eleven thirds. So the fourth derivative is minus eighty over eighty one one over the eleven. And I will need this because I said n equals four for some stupid reason. Eight times eleven is eight eight zero and three times eighty one is two forty three. X to the minus fourteen thirds. Fifth derivative at eight is some horrible number. Eight times ten over two will be three times one over three over fourteen. Okay, so what's my series? I guess I'm going back over here. So my series is then just take all of those numbers, divide them by the powers factorial, multiply by x minus eight to the n, right? So I'm only taking the first four terms, so I'm stopping here. It is two plus one twelve. Is the sign right? No, nobody told me that I got the sign wrong. All of the signs backwards. That's why nobody told me what's wrong because it was right. Okay, so this is a plus, that's a minus, this is a plus, that's a minus, this is a plus, this is a plus, that's a minus, this is a plus. Okay, good. Sorry, x minus two minus, the next term is, I have no idea what nine times thirty two is, two eighty eight, but I double put it over two so that's one forty four. That doesn't seem right. Two eighty eight, but then I have two on top so it's one forty four. So if my relationship is right, then I have to double it again. So let me just leave it. Over nine times thirty two times two factorial, x minus two squared, and then I ran out of room plus, so n equals four so I can do two more. Ten over twenty seven times two to the eighth, which is some number that I know but I don't know today, times three cubed, three factorial. Yeah, some number I know but I just don't know it today. X minus two cubed, and one more term, eighty over eighty one times two to the eleventh, which is another number I know but I don't know today. Four factorial, x minus four, that's a four. Okay, I'm getting screwed up. There we go. So this is my fourth degree polynomial approximation of the fourth group, which is way better than I need. So how good is this? Well, we have to look at, we have to look at what is the biggest this is over the region in question. This is an increasing function, right? This is an increasing function so it's biggest when x is ten. Smallest when x is six. So the bound that I want, so this is good over two forty three to the minus fourteen thirds divided by five factorial and then when we evaluate it at ten, that will be ten minus two to this thing. So this is some number you can punch in. Yeah. How did you know that it was good between that? Because of this theorem, which I've erased. So the thing that I erased says it's like the alternating series test, but it says the fifth derivative, the term you didn't use, except you take the x that makes it the biggest. Because I went to the fourth term here. Okay, so let's think about an alternating series. Maybe you remember this one. If I say I want to know how good, if I just take two terms of the alternating series, how good is that? How far off is that from the end? It's the value of the term you didn't use. This is the term I didn't use. So it's no worse than the term I didn't use. Yeah. Well, this is biggest in x equals ten. So the most error happens in x equals ten. Ten minus, why is that an eight? This should be an eight. All of these are eights. All these twos are eights. I have trouble telling two shall make parents. All of these twos are eights. So at six I get the same thing. Six minus eight. Ten minus eight. Absolute value. Oh, you're right. This should be a six. Woo. I'm sorry. Doesn't matter. I mean it doesn't matter. It's biggest at six. No, it's not. Yes it is. Right. One over, one over six is a fourteen over three. It's bigger than one over. It depends on the fourteen over. Because it's one over. But the bottom is bigger. This is smaller. So we do it at six. Not at ten. I'm sorry. It's a decreasing function. And so I'm sorry I screwed up. I always screwed up. That's what I do. It's my job. So, again, the way this works, it's actually very simple. I know it seems confusing. I'm sorry I made so many mistakes today. Is we calculate the Taylor series. So here the center is at eight. Even though I kept writing two. The center is at eight. So we calculate the Taylor series. We stop somewhere. And the error is given by the next term. But since the next term involves an x. So we can put an x here if you want. Since the next term involves an x. We have to say when it will be worse. And it's worse when this derivative is the biggest. Because we're looking at the error over a whole range. It's a function where we're saying how much is this function off from that function. And the difference will be controlled by the place where it works. So we have to look wherever this quantity is biggest. And I screwed up and put a ten where I should put a six. Usually it's at the middle or one of the two ends. Check the middle or the two ends. You really want to be careful like if you're doing an engineering project and your bridge will fall down if it's wrong. Then you should take the next derivative. Find out where it's zero. Determine whether the function is increasing or decreasing. Or has a maximum in the middle or whatever the hell it is. And use that. But usually it's at one end or the other or major in the middle. Okay? I want to move on. I'm tired of this stuff. I'm sure you are too. Are we okay with this? So you will have some problems on this. Feel free to ask if you're more confused. Let's move on to our next topic. So I know you wanted things to get easier so now we move on to the next topic. Yay! So our next topic... We went over binomial series. We went over binomial series last time. Well I can't spend a whole week on everything or we'll spend a year doing this class. Binomial series is just another series. It's the same joke. If you have a binomial series question I'm happy to do it. Okay? Absolutely. Find them a point series for the function one over the square root of four minus x. I did one almost exactly like this in class. I did a nine instead of a four. So after the four out you have one over a half. So I won't do it completely. That series is negative. So this is one half one minus x over four minus one half. Negative. Negative. It's negative. Whatever I'm saying. Now write the binomial series. Make the substitution u equals x over four. So you have one half one minus u to the minus one half. This is a binomial series. This is minus one half to two is n. U n, but my u is x over four. Okay? So let me tell you how the choose thing works. So the choose thing, if this is not a whole number or is not bigger than this, is just start taking subtractive. So this is, I guess I'll write it here. This is one half. So I just said I wasn't going to do it and now I just did it. Oh well. So the first term here is zero. The next term, so I get x and get one. And then the next term here is one half over four. And the next term is, it's a minus one half. Then the next term is minus a half minus one more over two. So you just keep subtracting one each time as you kick up the factorial in the bottom. Okay? This is not what I'm doing. So this is how these work. Are you okay with this? You just manipulate it into another series and there it is. Okay. So I want to move on to a different topic. So I think this is appendix h. I don't know. Complex numbers for a little bit for 20 minutes. I'll do a little more in a minute. So how many of you use complex numbers a little bit in high school education? How many people have never used complex numbers? I think you're lying. Okay. So you may remember that sometimes, like when you're solving a quadratic equation or something, you have something like the square root of negative eight that shows up in there. So we can factor the eight out. Oh, that was bad. Sorry. How about the square root of negative 81? We can factor the 81 out and get nine. But we can't do anything with this square root of one. So we just call it a new number, which we call i. We started doing this in approximately 16 something, I think maybe even the 16th century, so maybe 15 something. This seems so unusual to them that they call these imaginary numbers. They're not really imagined well. They're just an extension of numbers. But they turn out to be very useful even in doing things where the solutions don't involve imaginary numbers. They're complex numbers, which we'll say in a second. So we just, for argument's sake for now, we define a new number whose square root is the square root of negative one. So that means that if we square this number, we get negative one. And we can extend the ordinary real numbers that we dealt with by adding these things on. The complex number is just an ordinary real number. It's a pair of things. So an ordinary real number plus some multiple of one of those things. And they turn out to be useful. We will hopefully get to some of the useful applications of it soon. But they will also come up later in the course. And they certainly come up in all sorts of applications. So if we have a number before a plus b i, so this is a real number. And that's a real number. This is called the real part, the complex number, a. And b is called the imaginary part. And usually when we have variables, which we will get to very soon, usually for real numbers we use the letter x or y to represent a real variable. And typically people use z for complex numbers. It's no hard and fast rule. You could use any letter. You could use the x. But often when you see f of z, they're trying to tell you this is a complex function. So we can do arithmetic with these things. And they work just like binomials. So binomials are two things. So if we add two of them together. So if we have three plus two i plus five minus seven i. We just add the real parts, add the imaginary parts. No surprise there. If we multiply these two things together, we multiply them like binomials if you agree to. So we foil it out. Three times five is 15. Three times minus seven is minus 21. I. Two i times five is ten i. And two i times minus seven i is minus 14 i squared. But that's actually a plus 14. Because i squared is minus one. So this collapses to 15 plus 14 minus, no, plus, minus 21 plus ten i. So that's 29 minus 11 i. So this is all really exciting. We can divide them two. If you divide, then you just rationalize the denominator to get the i's on top. So if I wanted to do, I don't know, one over one plus i is six. Or about let's do one minus i over one plus i. So if I want to do this division, then I do the trick where I want to turn this into something not involving an i. So I multiply by its conjugate, top and bottom. So on the bottom, one plus i times one minus i is one times one. The middle terms drop out. The middle terms minus i. I have one plus i quantity squared, which gives me, so one plus minus one is zero. So one times one is one. And then the middle term, I get minus i minus i. And then this term gives me plus i squared. Yeah, that's right. So this gives me minus two i over two, which is minus i, plus i squared. Okay. So that's all well and good and not terribly exciting. I want to actually interpret these in a slightly different way. Oh, by the way, I neglected to say this. When we have a complex number, I guess here. So if I have a complex number z equals a plus ib, its complex conjugate z bar is a minus i. This is called the conjugate. I want to interpret this stuff geometrically. These complex numbers, a plus ib, well, they have two parts. So just like for real numbers, I can find it on the number line. So if I want to put, say, the real number five, that would be right here on the number line. I can have a number playing for complex numbers. They won't sit on the line because they have two parts. One plus i, where would I put one plus i on the line? It doesn't go on the line. So I can think of this as the complex. So in the complex plane, you put the real numbers along the x-axis and I put the imaginary numbers vertically along the y-axis. And if I have a point like one plus two i, I would plot it at four minutes. This seems stupid, but actually there's some nice geometrical intuition that goes with this. So I can interpret this as a vector. And this addition process, when I add two complex numbers like A plus iB plus C plus iB, this gives me A plus C from B plus C, which corresponds exactly to, and so one plus two i, let's put A plus iB here and let's put it C plus iB here. This corresponds exactly to an addition of vectors like that. So dealt with vectors in physics or in other contexts, addition of complex numbers is exactly the same as addition of vectors in the plane. Subtraction of complex numbers works exactly like subtraction of vectors. So, there's a lot of words that we're about to say. So adding and subtracting two numbers we can interpret this operation geometrically as addition and subtraction of vectors. Now vectors have another property. How many of you have dealt with vectors before? Okay, mostly everybody. Vectors have another way of thinking of them. They have a magnitude and they have a direction. So to these complex numbers we can assign a magnitude and a direction. So complex numbers also have magnitude and direction because we have this association between them. So if I have a complex number, if I have a complex number Z, which is A plus iB, its magnitude is written as absolute value and this is sometimes called the norm. It's not the guy from tears but it's norm of Z which is the square root of A squared plus B squared which if I look at this picture is this length. Well, this is the R in polar coordinates. It's called the argument of Z is the angle. If you were to think of Z or A plus iB in polar coordinates we can calculate the argument and the norm for arbitrary complex numbers in the same way that if you just convert the number to polar coordinates you get the same thing. How many people have seen this before? Fewer than some. So we can think of these complex numbers in polar coordinates or in rectangular coordinates and in fact we can switch back and forth. Now when we add complex numbers this is addition of vectors. There's no such thing as multiplication of vectors. It doesn't really make sense in terms of vectors but it does make sense in terms of complex numbers. So let's think about what this multiplication of vectors would do. So let's just write it out. So if I want to multiply well, when I'm multiplying A plus iB plus C plus iB then I multiply together the two real parts and then the two real parts and I will subtract off the product of the imaginary parts. That's the real part and then the resulting imaginary part is just the product of the two imaginary parts BC plus B. This is the same formula that I've written down well I did by example in there but suppose we think of this in polar coordinates. So I'm going to write A plus iB and C plus iB in polar coordinates instead right here. So this is R times the cosine of theta and this is R times the sine of theta. So this is the same A plus iB but I'm just writing it in terms of polar coordinates. So here I have R cosine theta plus sine theta and that's A plus iB and C plus iB is some other so this will be say s cosine of phi is s sine of phi. So phi is some other angle because I multiply two different numbers with different angles and different arguments together and so when I multiply them together well I guess it's already written there so then the product this times this so this is RS cosine of theta cosine of phi so that's this part here minus well BD is RS sine of theta sine of theta. Maybe this looks familiar to you maybe not. And then this part BC AD almost no time. So BC is RS sine theta sine What did I do wrong here? Yeah this is cosine and then the other one sine phi cosine theta So what is this? This is RS cosine of theta plus phi RS sine of theta plus phi In other words multiplying in complex numbers means splits and add the angles. When I multiply these two complex vectors together I want to multiply this complex number times this complex number What I do is I just take their links and multiply them together